33. s영역에서의 회로해석-마디, 메쉬해석, 중첩원리, 전원변환, 테브난, 노턴 이론

33. s영역에서의 회로해석-마디, 메쉬해석, 중첩원리, 전원변환, 테브난, 노턴 이론

s영역에서의 메쉬해석

$\displaystyle\frac{4}{\mathbf{s}+2}=\frac{3}{\mathbf{s}}\mathbf{I}_{1}+10(\mathbf{I}_{1}-\mathbf{I}_{2})$, $\displaystyle\frac{2}{\mathbf{s}+1}=10(\mathbf{I}_{2}-\mathbf{I}_{1})+4\mathbf{s}\mathbf{I}_{2}$

$\displaystyle\frac{4}{\mathbf{s}+2}=\left(10+\frac{3}{\mathbf{s}}\right)\mathbf{I}_{1}-10\mathbf{I}_{2}$, $\displaystyle\frac{2}{\mathbf{s}+1}=-10\mathbf{I}_{1}+(4\mathbf{s}+10)\mathbf{I}_{2}$

$\displaystyle\mathbf{I}_{1}=\frac{2\mathbf{s}(4\mathbf{s}^{2}+19\mathbf{s}+20)}{20\mathbf{s}^{4}+66\mathbf{s}^{3}+73\mathbf{s}^{2}+57\mathbf{s}+30}$, $\displaystyle\mathbf{I}_{2}=\frac{30\mathbf{s}^{2}+43\mathbf{s}+6}{(\mathbf{s}+2)(20\mathbf{s}^{3}+26\mathbf{s}^{2}+21\mathbf{s}+15)}$

$i_{1}(t)=\mathcal{L}^{-1}\{\mathbf{I}_{1}(\mathbf{s})\}=-96.39e^{-2t}-344.8e^{-t}+841.2e^{-0.15t}\cos0.8529t+199.7e^{-0.15t}\sin0.8529t\text{mA}$ $i_{2}(t)=\mathcal{L}^{-1}\{\mathbf{I}_{2}(t)\}=-481.9e^{-2t}-241.4e^{-t}+723.3e^{-0.15t}\cos0.8529t+472e^{-0.15t}\sin0.8529t\text{mA}$

s영역에서의 마디해석($v_{C}(0^{-})=2\text{V}$)

$\displaystyle-1=\frac{\mathbf{V}_{x}-7/\mathbf{s}}{2/\mathbf{s}}+\mathbf{V}_{x}+\frac{\mathbf{V}_{x}-4/\mathbf{s}}{4\mathbf{s}}$

$\displaystyle\mathbf{V}_{x}=\frac{10\mathbf{s}^{2}+4}{\mathbf{s}(2\mathbf{s}^{2}+4\mathbf{s}+1)}=\frac{5\mathbf{s}^{2}+2}{\mathbf{s}(\mathbf{s}+1+\sqrt{2}/2)(\mathbf{s}+1-\sqrt{2}/2)}$

$v_{x}(t)=\mathcal{L}^{-1}\{\mathbf{V}_{x}\}=4+6.864e^{-1.707t}-5.864e^{-0.2929t}\text{V}$

$\displaystyle\mathbf{V}_{C}=\frac{7}{\mathbf{s}}-\mathbf{V}_{x}=\frac{4\mathbf{s}^{2}+28\mathbf{s}+3}{\mathbf{s}(2\mathbf{s}^{2}+4\mathbf{s}+1)}$, $\displaystyle v_{C}(0^{+})=\lim_{\mathbf{s}\,\rightarrow\,\infty}{\mathbf{s}\mathbf{V}_{c}}=2\text{V}$.

s영역에서의 전원변환

$\displaystyle\mathbf{Z}_{1}=\frac{2}{\mathbf{s}}||10=\frac{20}{10\mathbf{s}+2}$

$\displaystyle\mathbf{Z}_{2}=\mathbf{Z}_{1}+\frac{2}{\mathbf{s}}=\frac{40\mathbf{s}+4}{\mathbf{s}(10\mathbf{s}+2)}$

$\displaystyle\mathbf{V}_{2}(\mathbf{s})=\frac{\mathbf{s}^{2}}{\mathbf{s}^{2}+9}\frac{20}{10\mathbf{s}+2}$

$\displaystyle\mathbf{V}(\mathbf{s})\frac{9\mathbf{s}}{9\mathbf{s}+(40\mathbf{s}+4)/\mathbf{s}(10\mathbf{s}+2)}=\frac{180\mathbf{s}^{4}}{(\mathbf{s}^{2}+9)(90\mathbf{s}^{3}+18\mathbf{s}^{2}+40\mathbf{s}+4)}$

$\displaystyle v(t)=\mathcal{L}^{-1}\{\mathbf{V}(\mathbf{s})\}=5.590\times10^{-5}e^{-0.1023t}+2.098\cos(3t+3.912^{\circ})+0.1017e^{-0.04885t}\cos(0.6573t+157.9^{\circ})\text{V}$.

s영역에서의 테브난 등가회로

입력 임피던스(1$\text{A}$의 시험전원 연결): $\mathbf{Z}_{in}=\frac{\mathbf{V}_{\text{in}}}{1\text{A}}=\mathbf{V}_{\text{in}}$

$\displaystyle1+g\mathbf{V}_{\pi}=\frac{\mathbf{V}_{\text{in}}}{\mathbf{Z}_{\text{eq}}}$, $\displaystyle\mathbf{Z}_{eq}=R_{E}||\frac{1}{\mathbf{s}}C_{\pi}||r_{\pi}=\frac{R_{E}r_{\pi}}{r_{\pi}+R_{E}+\mathbf{s}R_{E}r_{\pi}C_{\pi}}$

$\mathbf{V}_{\pi}=-\mathbf{V}_{\text{in}}$, $\mathbf{Z}_{\text{in}}=\mathbf{V}_{\text{in}}$, $\displaystyle1-g\mathbf{V}_{\text{in}}=\frac{\mathbf{V}_{\text{in}}}{\mathbf{Z}_{\text{eq}}}$, $\displaystyle\mathbf{V}_{\text{in}}=\frac{\mathbf{Z}_{eq}}{1+g\mathbf{Z}_{eq}}=\frac{1}{1/\mathbf{Z}_{eq}+g}$, $\displaystyle\mathbf{Z}_{\text{in}}=\frac{R_{E}r_{\pi}}{r_{\pi}+R_{E}+\mathbf{s}R_{E}r_{\pi}C_{\pi}+gR_{E}r_{\pi}}$.

참고자료:

Engineering Circuit Analysis 8th edition, Hayt, Kemmerly, Durbin, McGraw-Hill